3.34.82 \(\int \frac {-e^x x+e^{12+4 x} x^3 (3+2 x)}{x+2 e^{12+3 x} x^4+e^{24+6 x} x^7} \, dx\)

Optimal. Leaf size=19 \[ \frac {e^x}{-1-e^{3 (4+x+\log (x))}} \]

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Rubi [F]  time = 3.79, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-e^x x+e^{12+4 x} x^3 (3+2 x)}{x+2 e^{12+3 x} x^4+e^{24+6 x} x^7} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-(E^x*x) + E^(12 + 4*x)*x^3*(3 + 2*x))/(x + 2*E^(12 + 3*x)*x^4 + E^(24 + 6*x)*x^7),x]

[Out]

-1/3*Defer[Int][E^x/(1 + E^(4 + x)*x)^2, x] - Defer[Int][E^x/(x*(1 + E^(4 + x)*x)^2), x]/3 + Defer[Int][E^x/(x
*(1 + E^(4 + x)*x)), x]/3 - Defer[Int][E^x/(1 - E^(4 + x)*x + E^(8 + 2*x)*x^2)^2, x] + Defer[Int][E^(4 + 2*x)/
(1 - E^(4 + x)*x + E^(8 + 2*x)*x^2)^2, x] - Defer[Int][E^x/(x*(1 - E^(4 + x)*x + E^(8 + 2*x)*x^2)^2), x] + Def
er[Int][(E^(4 + 2*x)*x)/(1 - E^(4 + x)*x + E^(8 + 2*x)*x^2)^2, x] + Defer[Int][E^x/(1 - E^(4 + x)*x + E^(8 + 2
*x)*x^2), x]/3 - Defer[Int][E^(4 + 2*x)/(1 - E^(4 + x)*x + E^(8 + 2*x)*x^2), x]/3 + Defer[Int][E^x/(x*(1 - E^(
4 + x)*x + E^(8 + 2*x)*x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (-1+e^{3 (4+x)} x^2 (3+2 x)\right )}{\left (1+e^{3 (4+x)} x^3\right )^2} \, dx\\ &=\int \left (-\frac {e^x (1+x)}{3 x \left (1+e^{4+x} x\right )^2}+\frac {e^x}{3 x \left (1+e^{4+x} x\right )}+\frac {e^x (1+x) \left (-1+e^{4+x} x\right )}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}-\frac {e^x \left (-3-x+e^{4+x} x\right )}{3 x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {e^x (1+x)}{x \left (1+e^{4+x} x\right )^2} \, dx\right )+\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )} \, dx-\frac {1}{3} \int \frac {e^x \left (-3-x+e^{4+x} x\right )}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )} \, dx+\int \frac {e^x (1+x) \left (-1+e^{4+x} x\right )}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx\\ &=\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )} \, dx-\frac {1}{3} \int \left (\frac {e^x}{\left (1+e^{4+x} x\right )^2}+\frac {e^x}{x \left (1+e^{4+x} x\right )^2}\right ) \, dx-\frac {1}{3} \int \left (-\frac {e^x}{1-e^{4+x} x+e^{8+2 x} x^2}+\frac {e^{4+2 x}}{1-e^{4+x} x+e^{8+2 x} x^2}-\frac {3 e^x}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )}\right ) \, dx+\int \left (\frac {e^x \left (-1+e^{4+x} x\right )}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}+\frac {e^x \left (-1+e^{4+x} x\right )}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {e^x}{\left (1+e^{4+x} x\right )^2} \, dx\right )-\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )^2} \, dx+\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )} \, dx+\frac {1}{3} \int \frac {e^x}{1-e^{4+x} x+e^{8+2 x} x^2} \, dx-\frac {1}{3} \int \frac {e^{4+2 x}}{1-e^{4+x} x+e^{8+2 x} x^2} \, dx+\int \frac {e^x \left (-1+e^{4+x} x\right )}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx+\int \frac {e^x \left (-1+e^{4+x} x\right )}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx+\int \frac {e^x}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )} \, dx\\ &=-\left (\frac {1}{3} \int \frac {e^x}{\left (1+e^{4+x} x\right )^2} \, dx\right )-\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )^2} \, dx+\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )} \, dx+\frac {1}{3} \int \frac {e^x}{1-e^{4+x} x+e^{8+2 x} x^2} \, dx-\frac {1}{3} \int \frac {e^{4+2 x}}{1-e^{4+x} x+e^{8+2 x} x^2} \, dx+\int \frac {e^x}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )} \, dx+\int \left (\frac {e^{4+2 x}}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}-\frac {e^x}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}\right ) \, dx+\int \left (-\frac {e^x}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}+\frac {e^{4+2 x} x}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {e^x}{\left (1+e^{4+x} x\right )^2} \, dx\right )-\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )^2} \, dx+\frac {1}{3} \int \frac {e^x}{x \left (1+e^{4+x} x\right )} \, dx+\frac {1}{3} \int \frac {e^x}{1-e^{4+x} x+e^{8+2 x} x^2} \, dx-\frac {1}{3} \int \frac {e^{4+2 x}}{1-e^{4+x} x+e^{8+2 x} x^2} \, dx-\int \frac {e^x}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx+\int \frac {e^{4+2 x}}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx-\int \frac {e^x}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx+\int \frac {e^{4+2 x} x}{\left (1-e^{4+x} x+e^{8+2 x} x^2\right )^2} \, dx+\int \frac {e^x}{x \left (1-e^{4+x} x+e^{8+2 x} x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 20, normalized size = 1.05 \begin {gather*} -\frac {e^x}{1+e^{12+3 x} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^x*x) + E^(12 + 4*x)*x^3*(3 + 2*x))/(x + 2*E^(12 + 3*x)*x^4 + E^(24 + 6*x)*x^7),x]

[Out]

-(E^x/(1 + E^(12 + 3*x)*x^3))

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fricas [A]  time = 0.71, size = 18, normalized size = 0.95 \begin {gather*} -\frac {e^{x}}{x^{3} e^{\left (3 \, x + 12\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*exp(x)*exp(3*log(x)+3*x+12)-exp(x)*x)/(x*exp(3*log(x)+3*x+12)^2+2*x*exp(3*log(x)+3*x+12)+x)
,x, algorithm="fricas")

[Out]

-e^x/(x^3*e^(3*x + 12) + 1)

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giac [A]  time = 0.13, size = 18, normalized size = 0.95 \begin {gather*} -\frac {2 \, e^{x}}{x^{3} e^{\left (3 \, x + 12\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*exp(x)*exp(3*log(x)+3*x+12)-exp(x)*x)/(x*exp(3*log(x)+3*x+12)^2+2*x*exp(3*log(x)+3*x+12)+x)
,x, algorithm="giac")

[Out]

-2*e^x/(x^3*e^(3*x + 12) + 1)

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maple [A]  time = 0.04, size = 19, normalized size = 1.00

method	result	size
risch	\(-\frac {{\mathrm e}^{x}}{x^{3} {\mathrm e}^{3 x +12}+1}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+3)*exp(x)*exp(3*ln(x)+3*x+12)-exp(x)*x)/(x*exp(3*ln(x)+3*x+12)^2+2*x*exp(3*ln(x)+3*x+12)+x),x,method
=_RETURNVERBOSE)

[Out]

-exp(x)/(x^3*exp(3*x+12)+1)

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maxima [A]  time = 0.44, size = 18, normalized size = 0.95 \begin {gather*} -\frac {e^{x}}{x^{3} e^{\left (3 \, x + 12\right )} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*exp(x)*exp(3*log(x)+3*x+12)-exp(x)*x)/(x*exp(3*log(x)+3*x+12)^2+2*x*exp(3*log(x)+3*x+12)+x)
,x, algorithm="maxima")

[Out]

-e^x/(x^3*e^(3*x + 12) + 1)

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mupad [B]  time = 2.40, size = 18, normalized size = 0.95 \begin {gather*} -\frac {{\mathrm {e}}^x}{x^3\,{\mathrm {e}}^{3\,x+12}+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*exp(x) - exp(x)*exp(3*x + 3*log(x) + 12)*(2*x + 3))/(x + 2*x*exp(3*x + 3*log(x) + 12) + x*exp(6*x + 6*
log(x) + 24)),x)

[Out]

-exp(x)/(x^3*exp(3*x + 12) + 1)

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sympy [A]  time = 0.16, size = 17, normalized size = 0.89 \begin {gather*} - \frac {e^{x}}{x^{3} e^{12} e^{3 x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+3)*exp(x)*exp(3*ln(x)+3*x+12)-exp(x)*x)/(x*exp(3*ln(x)+3*x+12)**2+2*x*exp(3*ln(x)+3*x+12)+x),x
)

[Out]

-exp(x)/(x**3*exp(12)*exp(3*x) + 1)

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